SECTION Electrostatics This sectio, based o Chapter of Griffiths, covers effects of electric fields ad forces i static (timeidepedet) situatios. The topics are: Electric field Gauss s Law Electric potetial Work ad eergy Coductors Superpositio of forces Q Electric field Suppose we have some static arragemet of source charges (,, 3 ad so o), ad we wat to be able to determie the force o aother charge (a test charge Q). The priciple of superpositio states that the iteractio of two charges is idepedet of ay other charges preset. i To fid the force from, we ca igore the other charges. The we ca do the same foach of the other charges, ad do a vector sum of the forces. F = F F F 3 Hece the problem is reduced to solvig for the force betwee ay oe source charge ad the test charge. Coulomb s Law Now we eed the force that that ay source charge exerts o our test charge Q. From experimets, it s bee determied that Q r r origi The costat ε 0 is called the permittivity of free space. I SI uits: 0 = 8.850 C N m The separatio vector betwee ad Q is as defied earlier: = r r The force is attractive if ad Q have opposite sigs, ad repulsive if they have the same sig. Electric Field Now goig back to the case where there are may source charges, the total force is: F = F F = Q ˆr e Q $ ˆr e & = Q % r ˆr e e r ˆr $ e & e % We ca ow defie the electric field vector E so that F = Q E, so that E(r) = i r ˆr ei i= ei for a system of charges. The electric field is a fuctio of positio, but does ot deped o the test charge. It is a real etity, because we will show that it carries eergy ad mometum. 3
Cotiuous charge distributios By chagig the sum to a itegral, we ca calculate the field due to a cotiuous distributio of charge: E(r) = d ˆ Here d is a elemet of charge ad its value will deped o what shape we are itegratig over. There are 3 cases of iterest: Lie charge (λ per uit legth) d d l d l E(r) = ( r ) ˆr e d l lie P Surface charge (σ per uit area) d d a d a E(r) = ( r ) ˆr e d a surface P Volume charge (ρ per uit volume) d P d d E(r) = ( r ) ˆr e d $ volume Gauss s Law Electric field lies are a useful way to represet the electric field schematically. They origiate at positive charges, termiate o egative charges, ad ca exted to ifiity. The deser the lies, the higher the electric field. Electric Flux By cosiderig ay surface i a electric field, we ca calculate the flux, which is a measure of the umber of field lies crossig a surface (of area A held at right agles): 4
Flux = EA Flux = EA cos θ Flux = 0 I geeral, E = E da surface Schematically it is expected that a closed surface which does ot eclose ay charge has a eual umber of lies eterig ad leavig (zero et flux). Foxample (see opposite): If, o the other had, a surface ecloses a charge, there will be a et umber of lies eterig or leavig, i.e., a et flux that is proportioal to the charge. eclosed. From these ualitative argumets, we ca coclude that the flux through a closed surface depeds oly o the amout of charge eclosed by the surface. Gauss s Law As a simple example, the flux through a sphere, with radius r, cetred o a charge placed at the origi is: E da = sphere r 4r = Although the math is easier with a sphere, it works for ay shape, ad the charge ca be positioed aywhere. A geeral proof (usig solid agles) goes as follows: 0 r da Solid agle dω E Cosider a poit charge aywhere withi a arbitrary shaped volume, ad deote by da a vectolemet of surface area (meaig its magitude is the area da ad its directio is the outward ormal to the surface). The directio of the electric field E is radially outward from the charge. Flux through the elemet of area where the last step follows from the defiitio of solid agle. Therefore the total flux for the closed surface is E da = d = surface alldirectios 0 5
I geeral for charges eclosed, by the priciple of superpositio E = E ad so E da i = E i da = i= We ca ow summarize this as follows. Gauss s Law says that the flux of E through a closed surface is: where the total charge eclosed is Q ec. (The charges outside make o differece to the result.) This is really just aother way of statig Coulomb s Law (i a itegral format), but it is a very useful result for cases which have spherical, cylidrical ad plae symmetry (because the itegrals the become easy). Gauss s Law i differetial form We ca chage the itegral form of Gauss s Law to a euivalet differetial form (which is eater, though the itegral form is more geerally useful). Sice this is true o matter what volume we choose, it follows that: Gauss s Law applicatios E da = surface Q ec 0 i= ( ) First, we apply the divergece theorem to the lefthad side: E da = Ed surface volume The, if we have a cotiuous charge distributio, the righthad side becomes: Q ec = d volume 0 0 Spherical symmetry: To fid the field at radius r, choose a surface with that radius. E = 0 i= $ ' & i ) % 0 ( Cylidrical symmetry: Agai, choose a surface (this time cylidrical) with radius s. Plae symmetry: Choose a flat pillbox, straddlig the surface. The electric field must be costat o that surface ad perpedicular to it, so that the surface itegral becomes a multiplicatio. Examples will be doe i classes. 6
Electric potetial Curl of E Suppose we wat to fid the curl of the field E due to a sigle poit charge at the origi: E = r ˆr We could do this directly, but aother way is to calculate the itegral of E alog some lie: dl = dr ˆr rd I spherical coordiates: ˆ rsid ˆ b b b $ ' * E dl = dr = a a So: r % & r ( ) = a a, / b. E dl For a closed path, we have a = b, ad = 0 Fially, by Stokes theorem (see Sectio ): Electric Potetial We saw i Sectio that if the curl of a vector field is 0, it ca be represeted as the gradiet of a scalar. Sice we kow the curl of a electrostatic field is always 0, we ca represet it as the gradiet of a scalar field, called the electric potetial V. Beig a scalar, the potetial is usually much easier to work with tha the field vector, which has compoets. It is ofte easier to work out the potetial due to a charge distributio first, ad the calculate the field usig: E = V Notes: The origi of V is arbitrary because we ca add ay costat to V with chagig E. Therefore we choose whatever is coveiet. Uless the charge distributio is ifiite, we usually take V = 0 at ifiity. The electric potetial, like the electric field, obeys the superpositio priciple. The electric potetial has uits of Volts. Poisso s Euatio ad Laplace s Euatio The fudametal electric field euatios, E = ad E = 0, 0 ca be rewritte i terms of the potetial V. For the divergece of electric field: E = (V ) = V E = 0 So, from Gauss s Law: V = 0 This is called Poisso s euatio. I the special case of ρ = 0 (zero charge desity i the regio), we have: V = 0 This is kow as Laplace s euatio. We will look at solutios to Poisso s ad Laplace s euatio i Sectio 3. I may problems we are give a charge distributio ad we wat to fid the electric potetial V. We ca (i priciple) itegrate Poisso s euatio to get V. First, for a poit charge at the origi: 7
r V(r) = r d r = % ( $ & ' r ) * = r I geeral, for a poit charge at ay positio: V(r) = Ad we ca write dow the potetial for may charges usig the superpositio priciple. V(r) = i i= For a cotiuous (volume) distributio of charge, we ca itegrate as: ( r ) V(r) = d $ Summary of relatioships i r V(r) = ( r ) d $ E = 0 V = 0 E = 0 Discotiuities ad boudary coditios E above V V(r) = E = V E dl E A The electric field chages discotiuously at a sheet of charge. It will be useful to determie how the field chages at this boudary. Usig Gauss s Law o the little pillbox, we get: E below If we make the thickess ε extremely small, the sides of the box do ot cotribute: E above E below = 0 So we have a boudary coditio that the perpedicular compoet of the field is discotiuous by a amout σ / ε 0 at ay boudary that has some surface charge. 8
l E above E below Next we look at the tagetial compoet of E. Take a small loop very close to the surface. We kow that from Stokes Law. Now we ca make the vertical legth ε very small (compared to l), so that the sides do ot cotribute to the itegral. This meas that the top ad bottom must cacel, so E above E below = 0 We ca combie these two boudary coditios i vector form as: E above E below = ˆ 0 a b Kowig the chage i electric field, we ca deduce the chage i potetial V: If we make poit a very close to b, the itegral will shrik to 0, so: Of course, we kow that the gradiet of V still has a discotiuity (because the gradiet of V is just E ad we have show that the perpedicular compoet of E has a discotiuity). Work to move a charge Work ad eergy What is the work eeded to move charge Q from poit a to poit b i the presece of ay distributio of charges? At ay poit o the path, the electric force o Q is b Q The work doe (by you) i movig the charge is therefore: i Obviously, the work is idepedet of path: this is why we ca call the electrostatic force coservative. The expressio for the work doe ca be rearraged as: V(b)V(a) = W Q We see that the differece i electric potetial is the work per uit charge to move a particle betwee the poits. To brig a charge from far away, the work is: 9
So, if we choose to set V = 0 at ifiity, Hece potetial is just potetial eergy per uit charge (like the field is force per uit charge). Eergy of a charge distributio 3 W = Q[V(r)V()] W = QV(r) Now we ca look at the work eeded to assemble a distributio of poit charges. Imagie doig this by brigig them i from ifiity oe by oe. No work is eeded to brig the first charge from ifiity (W = 0). 3 3 The work to brig i the d charge is: Ad the a 3 rd charge: W 3 = 3 $ & 3 3 % Ad the a 4 th charge: W 4 = 4 $ 3 &, assumig the case of 4 charges i total. 4 4 % We add these separate amouts up to get the total work doe to brig these four charges from ifiity: W = 3 4 3 4 $ 3 4 & 3 4 3 4 34 % For charges, this geeralizes to become: The factor of ½ is to avoid double coutig each term, ad obviously we eed to exclude terms with i ad j the same. This is useful i this form, but it ca also be writte aother way by takig out the sum over i: W = & i j % r ( i= $ j=(i j) eij ' The term i the brackets is just the electric potetial V at the positio of the i th charge. We ca ow rewrite the expressio for the work as: Eergy of a cotiuous distributio We ca fid the eergy of a cotiuous charge distributio by chagig the sum i the last expressio to a itegral: W = V d This is itself ofte a useful expressio, but there s aother useful itegral form i terms of the electric field. First, we use Gauss s Law to get the charge desity i terms of E: 34 W = i j W = i= i V(r i ) i= j=(i j) ij = 0 E This implies W = 0 ( E) V d 0
We ca ext trasfer the derivative from the electric field to the electric potetial, by usig itegratio by parts: W = 0 % $ E V d $ V E da' & ( = 0 % $ E Ed $ V E da' & ( where i the last step we used the defiitio E = V. We ow have a volume itegral over a volume that must be large eough to cotai all the charge distributio, plus a surface itegral (surroudig that volume). However, if we exted the volume to all space, the secod itegral goes to 0 because E ad V go to zero fast eough. So we fid the eergy as: W = 0 E d allspace Note that the superpositio priciple does ot hold foergy. (This is because the eergy of two systems has cross terms i the fields): W tot = 0 E tot d = 0 (E E ) d = 0 Coductors Basic properties of coductors I a coductor, the charges move aroud freely. I metals, the electros carry charge: i liuids, positive ad egative ios do. For ow, we will look at perfect or ideal coductors, which have ulimited free charge ad o resistace. We wat to kow what happes to the electric field, the charge desity ad the potetial. Exteral E E = 0 Electric potetial i a coductor: Sice the electric field is zero iside the coductor, the chage i potetial is also zero. This meas that the coductor is a euipotetial. Electric field just outside a coductor: The field cacels iside the coductor; outside the field may be ozero. The exteral field must, however, be perpedicular to the surface at all poits (if there was a compoet alog the surface, charge would flow to eutralize it). Iduced charge o a coductor The eergy ca therefore be viewed as stored i the electric field. (E E E E )d = W W 0 E E d Electric field iside a coductor: Sice charge ca move freely, the charges iside a coductor will arrage themselves to cacel ay exteral field, so the electric field iside a coductor is always 0. Charge desity iside a coductor: Sice the electric field iside the coductor is zero, the et charge must also be zero (eual amouts of positive ad egative charge). Charge desity at the surface of a coductor: If the coductor is i a electric field, positive ad egative charges separate util the field is cacelled. These ubalaced charges must reside o the surface oly. A coductor i the presece of a charge will be attracted to the charge because of the charge iduced o the surface of the coductor. If we ow place a charge iside a cavity i a coductor, the ier surface will acuire the opposite charge () to cacel the field iside the coductor. The outer surface will have charge, so that the total et charge of the coductor plus the cavity is. If there is o charge i the cavity, the field i the cavity must be zero.
Surface charge o coductors We calculated earlier the chage i electric field at a surface charge. Sice the field iside a coductor is zero, it follows that the field just outside must be: E = 0 ˆ We ca also write the surface charge i terms of the potetial: = 0 V Force o a coductor The force o a charge is just the charge times the electric field. I the case of a surface charge, the field is discotiuous: the force o the surface charge is the local surface charge times the average of the field iside ad outside the surface. For a uit area: f = E average = (E E ) above below For a coductor, the, the force per uit area is: f = ˆ 0 where the derivative is i the directio of the ormal. which pulls the coductor ito the field. We ca also look at this as a pressure, ad we ca rewrite it i terms of the field just outside the coductor: Capacitors Q P = 0 E Imagie two coductors (of ay shape): oe has charge Q, the other charge Q. We ca look at the potetial differece betwee them (each, of course, must be a euipotetial): Q From Coulomb s Law, we kow that the field is proportioal to Q; from the above euatio, we see that V must also be proportioal to Q. The proportioality costat is called the capacitace C: C = Q V Work ad capacitors We wat the work doe to charge a capacitor from charge 0 iitially to charge Q fially. Imagie doig this gradually, so that at some itermediate stage the charge is (which meas the voltage differece is /C). The work to trasfer a small extra amout of charge (call it d) from the positive plate of a capacitor to the egative plate is foud from the euatio for work: dw = voltage d = C d We ca itegrate from = 0 to = Q to get the total work: Q W = C d = Q 0 C Sometimes it is more useful to express the result i terms of voltage: W = CV